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IceJOKER [234]
3 years ago
7

Explain the relationship between an individual and a Society​

Chemistry
1 answer:
Rufina [12.5K]3 years ago
6 0

Answer:

the relationship between an individual and a society is society doesn't exist independently without an individual .the individual lives and acts within society but society is nothing ,in spite of the combination of individual s for cooperation effort.

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What is the mole fraction of each component in a solution made by mixing 300 g of ethanol(C2H5OH) and 500 g of water?
Snowcat [4.5K]

Answer:

The mole fraction of ethanol and water in the solution are 0.19 and 0.81 respectively.

Explanation:

The Molar Fraction is a way of measuring the concentration that expresses the proportion in which a substance is found with respect to the total moles of the solution, which are calculated by adding the moles of solute (s) and solvent.

It is calculated by dividing the number of moles of one of the components by the total number of moles of the solution.

The sum of all the molar fractions of the substances present in a solution is equal to 1.

So, the expression to calculate the mole fraction is:

Molar fraction (Xi)=\frac{moles of sudstance}{total moles of the solution}

Being the molar mass of the compounds:

  • Ethanol 46 \frac{g}{mol}
  • Water 18 \frac{g}{mol}

the number of moles that represent the mass quantities is calculated as:

  • Moles of ethanol: 300 grams* \frac{1 mol}{46 grams} = 6.52 moles
  • Moles of water: 500 grams* \frac{1 mol}{18 grams} = 27.78 moles

So the total moles in solution are 6.52 moles + 27.78 moles = 34.3 moles

The mole fraction of ethanol in the solution is calculated as:

mole fraction of ethanol=x_{ethanol} =\frac{6.52 moles}{34.3 moles}

x_{ethanol} =0.19

The mole fraction of water in the solution is calculated as:

mole fraction of water=x_{water} =\frac{27.78 moles}{34.3 moles}

x_{water} =0.81

Then:

x_{water} +x_{ethanol} = 0.81 + 0.19

x_{water} +x_{ethanol} =1

<u><em>The mole fraction of ethanol and water in the solution are 0.19 and 0.81 respectively.</em></u>

 

.

8 0
3 years ago
What is manganese dioxide in preparation of oxygen?
Levart [38]

Answer:

When manganese(IV) oxide is added to hydrogen peroxide, bubbles of oxygen are given off. To make oxygen in the laboratory, hydrogen peroxide is poured into a conical flask containing some manganese(IV) oxide. The gas produced is collected in an upside-down gas jar filled with water.

5 0
3 years ago
A mixture of A and B is capable of being ignited only if the mole percent of A is 6 %. A mixture containing 9.0 mole% A in B flo
atroni [7]

Explanation:

As it is given that mixture (contains 9 mol % A and 91% B) and it is flowing at a rate of 800 kg/h.

Hence, calculate the molecular weight of the mixture as follows.

             Weight = 0.09 \times 16.04 + 0.91 \times 29

                          = 27.8336 g/mol

And, molar flow rate of air and mixture is calculated as follows.

                    \frac{800}{27.8336}

                     = 28.74 kmol/hr

Now, applying component balance as follows.

                0.09 \times 28.74 + 0 \times F_{B} = 0.06F_{p}

                   F_{p} = 43.11 kmol/hr

                   F_{A} + F_{B} = F_{p}

                          F_{B} = 43.11 - 28.74

                                      = 14.37 kmol/hr

So, mass flow rate of pure (B), is F_{B} = 14.37 \times 29

                                                    = 416.73 kg/hr

According to the product stream, 6 mol% A and 94 mol% B is there.

             Molecular weight of product stream = Mol. weight \times 43.11 kmol/hr

                                  = 0.06 \times 16.04 + 0.94 \times 29

                                  = 28.22 g/mol

Mass of product stream = 1216.67 kg/hr

Hence, mole of O_{2} into the product stream is as follows.

                    0.21 \times 0.94 \times 43.11

                      = 8.5099 kmol/hr \times 329 g/mol

                      = 272.31 kg/hr

Therefore, calculate the mass % of O_{2} into the stream as follows.

                 \frac{272.31}{1216.67} \times 100

                     = 22.38%

Thus, we can conclude that the required flow rate of B is 272.31 kg/hr and the percent by mass of O_{2} in the product gas is 22.38%.

4 0
3 years ago
How many moles are in 5g of NaOH
Lady bird [3.3K]
5g NaOH x 1 mol NaOH/ 39.99g NaOH = 0.125 mol NaOH 
7 0
4 years ago
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When an oceanic plate and continental plate collide, the oceanic plate __________.
Naya [18.7K]
The less buoyant oceanic plate sinks and slides under the continental plate
5 0
3 years ago
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